package 算法.leetcode.offer;

import java.util.Stack;

/**
 * 定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
 *
 *  
 *
 * 示例:
 *
 * MinStack minStack = new MinStack();
 * minStack.push(-2);
 * minStack.push(0);
 * minStack.push(-3);
 * minStack.min();   --> 返回 -3.
 * minStack.pop();
 * minStack.top();      --> 返回 0.
 * minStack.min();   --> 返回 -2.
 *
 * @author lchenglong
 * @date 2022/3/19
 */
public class Offer30 {
    class MinStack {

        Stack<Integer> stack ;
        Stack<Integer> minStack ;
        public MinStack() {
            stack = new Stack<Integer>();
            minStack = new Stack<Integer>();
        }

        public void push(int x) {
            stack.push(x);
            if (minStack.empty()||minStack.peek()>=x)
                minStack.push(x);

        }

        public void pop() {
            if (stack.pop().equals(minStack.peek()))
                minStack.pop();
        }

        public int top() {
            Integer pop = stack.peek();
            return pop;
        }

        public int min() {
            return minStack.peek();
        }
    }
}
